Math 752 Topology Lecture Notes Laurenţiu Maxim May 3, 2013 Contents 1 Selected topics in Homology 3 1.1 Cellular Homology . .3 1.1.1 Degrees . .3 1.1.2 How to Compute Degrees? . .6 1.1.3 CW Complexes . .8 1.1.4 Cellular Homology . .9 1.2 Euler Characteristic . 18 1.3 Lefschetz Fixed Point Theorem . 20 1.4 Homology with General Coefficients . 23 1.5 Universal Coefficient Theorem for Homology . 26 1.5.1 Tensor Products . 26 1.5.2 The Tor functor and the Universal Coefficient Theorem . 28 2 Basics of Cohomology 33 2.1 Cohomology of a chain complex: definition . 33 2.2 Relation between cohomology and homology . 34 2.2.1 Ext groups . 34 2.2.2 Universal Coefficient Theorem . 35 2.3 Cohomology of spaces . 36 2.3.1 Definition and immediate consequences . 36 2.3.2 Reduced cohomology groups . 38 2.3.3 Relative cohomology groups . 39 2.3.4 Induced homomorphisms . 40 2.3.5 Homotopy invariance . 40 2.3.6 Excision . 41 2.3.7 Mayer-Vietoris sequence . 42 2.3.8 Cellular cohomology . 43 3 Cup Product in Cohomology 49 3.1 Cup Products: definition, properties, examples . 49 3.2 Application: Borsuk-Ulam Theorem . 60 3.3 Künneth Formula . 64 3.3.1 Cross product . 64 1 3.3.2 Künneth theorem in cohomology. Examples . 65 3.3.3 Künneth exact sequence and applications . 70 4 Poincaré Duality 73 4.1 Introduction . 73 4.2 Manifolds. Orientation of manifolds . 74 4.3 Cohomolgy with Compact Support . 79 4.4 Cap Product and the Poincaré Duality Map . 82 4.5 The Poincaré Duality Theorem . 85 4.6 Immediate applications of Poincaré Duality . 90 4.7 Addendum to orientations of manifolds . 92 4.8 Cup product and Poincaré Duality . 97 4.9 Manifolds with boundary: Poincaré duality and applications . 100 4.9.1 Signature . 102 4.9.2 Connected Sums . 103 5 Basics of Homotopy Theory 105 5.1 Homotopy Groups . 105 5.2 Relative Homotopy Groups . 111 5.3 Homotopy Groups of Spheres . 114 5.4 Whitehead’s Theorem . 117 5.5 Fibrations and Fiber Bundles . 121 2 Chapter 1 Selected topics in Homology Note: Knowledge of simplicial and singular homology will be assumed. 1.1 Cellular Homology 1.1.1 Degrees Definition 1.1.1. The degree of continuous map f : Sn ! Sn is defined as: deg f = f∗(1) (1.1.1) n n where 1 2 Z denotes the generator, and f∗ : Hen(S ) = Z ! Hen(S ) = Z is the homomor- phism induced by f in homology. The degree has the following properties: 1. deg idSn = 1. Proof. This is because (idSn )∗ = id which is multiplication by the integer 1. 2. If f is not surjective, then deg f = 0. Proof. Indeed, suppose f is not surjective, then there is a y2 = Imagef. Then we can factor f in the following way: f Sn > Sn > g > h Sn n fyg n ∼ n n Since S −fyg = R which is contractible, Hn(S nfyg) = 0. Therefore f∗ = h∗g∗ = 0, so deg f = 0. 3 3. If f ∼= g, then deg f = deg g. Proof. This is because f∗ = g∗. Note that the converse is also true (by a theorem of Hopf). 4. deg(g ◦ f) = deg g · deg f. Proof. Indeed, we have that (g ◦ f)∗ = g∗ ◦ f∗. 5. If f is a homotopy equivalence (so there exists a g so that g◦f ' idSn ), then deg f = ±1. ∼ Proof. This follows directly from 1, 3, and 4 above, since f ◦ g = idSn implies that deg f · deg g = deg idSn = 1. 6. If r : Sn ! Sn is a reflection across some n-dimensional subspace of Rn+1, that is, r(x0; : : : xn) 7! (−x0; x1; : : : ; xn), then deg r = −1. Proof. Without loss of generality we can assume the subspace is Rn × f0g ⊂ Rn−1. n n n Choose a CW structure for S whose n-cells are given by ∆1 and ∆2 , the upper and lower hemispheres of Sn, attached by identifying their boundaries together in the n n n standard way. Then consider the generator of Hn(S ): [∆1 − ∆2 ]. The reflection map n n n n n n r maps the cycle ∆1 − ∆2 to ∆2 − ∆1 = −(∆1 − ∆2 ). So n n n n n n n n r∗([∆1 − ∆2 ]) = [∆2 − ∆1 ] = [−(∆1 − ∆2 )] = −1 · [∆1 − ∆2 ] so deg r = −1. 7. If a : Sn ! Sn is the antipodal map (x 7! −x), then deg a = (−1)n+1 Proof. Note that a is a composition of n+1 reflections, since there are n+1 coordinates in x, each getting mapped by an individual reflection. From 4 above we know that composition of maps leads to multiplication of degrees. 8. If f : Sn ! Sn and Sf : Sn+1 ! Sn+1 is the suspension of f then deg Sf = deg f. Proof. Recall that if f : X ! X is a continuos map and ΣX = X × [−1; 1]=(X × {−1g;X × f1g) denotes the suspension of X, then Sf := f × id[−1;1]= ∼, with the same equivalence as in ΣX. Note that ΣSn = Sn+1. The Suspension Theorem states that ∼ Hei(X) = Hei+1(ΣX): 4 This can be proved by using the Mayer-Vietoris sequence for the decomposition ΣX = C+X [X C−X; where C+X and C−X are the upper and lower cones of the suspension joined along their bases: ! Hen+1(C+X) ⊕ Hen+1(C−X) ! Hen+1(ΣX) ! Hen(X) ! Hen(C+X) ⊕ Hen(C−X) ! Since C+X and C−X are both contractible, the end groups in the above sequence are ∼ both zero. Thus, by exactness, we get Hei(X) = Hei+1(ΣX), as desired. n n n n n Let C+S denote the upper cone of ΣS . Note that the base of C+S is S ×f0g ⊂ ΣS . n n n n Our map f induces a map C+f :(C+S ;S ) ! (C+S ;S ) whose quotient is Sf. The n n long exact sequence of the pair (C+S ;S ) in homology gives the following commutative diagram: @ 0 > H (C Sn;Sn) ' H (C Sn=Sn) > H (Sn) > 0 ei+1 + ei+1 + ∼ ei (Sf)∗ f∗ _ _ @ H (Sn+1) > H (Sn) ei+1 ∼ ei n n ∼ n+1 Note that C+S =S = S so the boundary map @ at the top and bottom of the diagram are the same map. So by the commutativity of the diagram, since f∗ is defined by multiplication by some integer m, then (Sf)∗ is multiplication by the same integer m. n n Example 1.1.2. Consider the reflection map: rn : S ! S defined by (x0; : : : ; xn) 7! (−x0; x1; : : : ; xn). Since rn leaves x1; x2; : : : ; xn unchanged we can unsuspend one at a time to get deg rn = deg rn−1 = ··· = deg r0; i i 0 0 where ri : S ! S by (x0; x1; : : : ; xi) 7! (−x0; x1; : : : ; xi). So r0 : S ! S by 0 x0 7! −x0. Note that S is two points but in reduced homology we are only looking at one integer. Consider 0 0 0 ! He0(S ) ! H0(S ) −! Z ! 0 0 0 where He0(S ) = f(a; −a) j a 2 Zg, H0(S ) = Z ⊕ Z, and :(a; b) 7! a + b. Then 0 0 (r0)∗ : He0(S ) ! He0(S ) is given by (a; −a) 7! (−a; a) = (−1)(a; −a). So deg rn = −1. 9. If f : Sn ! Sn has no fixed points then deg f = (−1)n+1. Proof. Consider the above figure. Since f(x) 6= x, the segment (1−t)f(x)+t(−x) from −x to f(x) does not pass through the origin in Rn+1 so we can normalize to obtain a homotopy: (1 − t)f(x) + t(−x) g (x) := : Sn ! Sn: t j(1 − t)f(x) + t(−x)j 5 Note that this homotopy is well defined since (1 − t)f(x) − tx 6= 0 for any x 2 Sn and t 2 [0; 1], because f(x) 6= x for all x. Then gt is a homotopy from f to a, the antipodal map. Exercises 1. Let f : Sn ! Sn be a map of degree zero. Show that there exist points x; y 2 Sn with f(x) = x and f(y) = −y. 2. Let f : S2n ! S2n be a continuous map. Show that there is a point x 2 S2n so that either f(x) = x or f(x) = −x. 3. A map f : Sn ! Sn satisfying f(x) = f(−x) for all x is called an even map. Show that an even map has even degree, and this degree is in fact zero when n is even. When n is odd, show there exist even maps of any given even degree. 1.1.2 How to Compute Degrees? Assume f : Sn ! Sn is surjective, and that f has the property that there exists some n −1 −1 y 2 Image(S ) so that f (y) is a finite number of points, so f (y) = fx1; x2; : : : ; xmg. Let Ui be a neighborhood of xi so that all Ui’s get mapped to some neighborhood V of y. So f(Ui − xi) ⊂ V − y. We can choose the Ui to be disjoint. We can do this because f is continuous. Let fjUi : Ui ! V be the restriction of f to Ui. Then: 6 f∗ Hn(Ui;Ui − xi) −! Hn(V; V − y) ' (excision) ' (excision) n n n n Hn(S ;S − xi) Hn(S ;S − y) ' l.e.s.